Wed. Apr 24th, 2024

Koshe Dekhi 8.5 Class 9

Koshe Dekhi 8.5 Class 9

উৎপাদকে বিশ্লেষণ

1. নিচের বহুপদী সংখ্যামালাগুলিকে করি :

(i) \fn_cm {\color{Blue} \left ( a+b \right )^{2}-5a-5b+6}

সমাধানঃ 

\fn_cm \left ( a+b \right )^{2}-5a-5b+6

\fn_cm =\left ( a+b \right )^{2}-5\left (a+b \right )+6

\fn_cm =\left ( a+b \right )^{2}-3\left (a+b \right )-2\left (a+b \right )+6

=\left ( a+b \right )\left [ \left ( a+b \right )-3 \right ]-2\left [ \left ( a+b \right )-3 \right ]

=\left ( a+b-3 \right )\left ( a+b-2 \right )

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(ii) \fn_cm {\color{Blue} \left ( x+1 \right )\left ( x+2 \right )\left ( 3x-1 \right )\left ( 3x-4 \right )+12}

সমাধানঃ 

\fn_cm \left ( x+1 \right )\left ( x+2 \right )\left ( 3x-1 \right )\left ( 3x-4 \right )+12

=\left [ \left ( x+1 \right )\left ( 3x-1 \right ) \right ]\left [ \left ( x+2 \right )\left ( 3x-4 \right ) \right ]+12

=\left [ 3x^{2}-x+3x-1 \right ]\left [ 3x^{2}-4x+6x-8 \right ]+12

=\left [ 3x^{2}+2x-1 \right ]\left [ 3x^{2}+2x-8 \right ]+12

=\left ( a-1 \right )\left ( a-8 \right )+12    [ ধরি, {\color{Blue} 3x^{2}+2x=a]}

=a^{2}-8a-a+8+12

=a^{2}-9a+20

=a^{2}-5a-4a+20

=a\left ( a-5 \right )-4\left ( a-5 \right )

=\left ( a-5 \right )\left ( a-4 \right )

=\left ( 3x^{2}+2x-5 \right )\left ( 3x^{2}+2x-4 \right ) [ a এর মান বসিয়ে পাই ]

=\left ( 3x^{2}+5x-3x-5 \right )\left ( 3x^{2}+2x-4 \right )

=\left [x\left ( 3x+5 \right )-1\left ( 3x+5 \right ) \right ]\left ( 3x^{2}+2x-4 \right )

=\left ( 3x+5 \right )\left ( x-1 \right )\left ( 3x^{2}+2x-4 \right )

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(iii) \fn_cm {\color{Blue} x\left ( x^{2}-1 \right )\left ( x+2 \right )-8}

সমাধানঃ 

\fn_cm x\left ( x^{2}-1 \right )\left ( x+2 \right )-8

=x\left [\left ( x+1 \right )\left ( x-1 \right ) \right ]\left ( x+2 \right )-8

[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]

=x\left ( x+1 \right )\left ( x-1 \right ) \left ( x+2 \right )-8

=\left [x\left ( x+1 \right ) \right ]\left [\left ( x-1 \right ) \left ( x+2 \right ) \right ]-8

=\left ( x^{2}+x \right )\left ( x^{2}+2x-x-2 \right )-8

=\left ( x^{2}+x \right )\left ( x^{2}+x-2 \right )-8

=a\left ( a-2 \right )-8 [ ধরি, {\color{Blue} x^{2}+x=a} ]

=a^{2}-2a-8

=a^{2}-4a+2a-8

=a\left ( a-4 \right )+2\left ( a-4 \right )

=\left ( a+2 \right )\left ( a-4 \right )

=\left ( x^{2}+x+2 \right )\left ( x^{2}+x-4 \right ) [ a এর মান বসিয়ে পাই ]

উৎপাদকে বিশ্লেষণ

(iv) \fn_cm {\color{Blue} 7\left ( a^{2}+b^{2} \right )^{2}-15\left ( a^{4}-b^{4} \right )+8\left ( a^{2}-b^{2} \right )^{2}}

সমাধানঃ 

\fn_cm 7\left ( a^{2}+b^{2} \right )^{2}-15\left ( a^{4}-b^{4} \right )+8\left ( a^{2}-b^{2} \right )^{2}

=7\left ( a^{2}+b^{2} \right )^{2}-15\left [\left ( a^{2}+b^{2} \right )\left ( a^{2}-b^{2} \right ) \right ]+8\left ( a^{2}-b^{2} \right )^{2}

[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]

=7p^{2}-15pq+8q^{2}  

[ ধরি, {\color{Blue} \left ( a^{2}+b^{2} \right )=p,\left ( a^{2}-b^{2} \right )=q} ]

=7p^{2}-7pq-8pq+8q^{2}

=7p\left ( p-q \right )-8q\left ( p-q \right )

=\left ( p-q \right )\left ( 7p-8q \right )

=\left ( a^{2}+b^{2}-a^{2}+b^{2} \right )\left [ 7\left ( a^{2}+b^{2} \right )-8\left (a^{2}-b^{2} \right ) \right ] [ p ও q এর মান বসিয়ে পাই ]

=2b^{2}\left ( 7a^{2}+7b^{2}-8a^{2}+8b^{2} \right )

=2b^{2}\left ( 15b^{2}-a^{2} \right )

উৎপাদকে বিশ্লেষণ

(v) \fn_cm {\color{Blue} \left ( x^{2}-1 \right )^{2}+8x\left ( x^{2}+1 \right )+19x^{2}}

সমাধানঃ 

\fn_cm \left ( x^{2}-1 \right )^{2}+8x\left ( x^{2}+1 \right )+19x^{2}

\fn_cm =\left ( x^{2}+1 \right )^{2}-4x^{2}+8x\left ( x^{2}+1 \right )+19x^{2}

{\color{Blue} \left [ \because \left ( a-b \right )^{2}=\left ( a+b \right )^{2}-4ab \right ]}

\fn_cm =\left ( x^{2}+1 \right )^{2}+8x\left ( x^{2}+1 \right )+15x^{2}

\fn_cm =\left ( x^{2}+1 \right )^{2}+5x\left ( x^{2}+1 \right )+3x\left ( x^{2}+1 \right )+15x^{2}

\fn_cm =\left ( x^{2}+1 \right )\left [\left ( x^{2}+1 \right )+5x \right ]+3x\left [\left ( x^{2}+1 \right )+5x \right ]

=\left ( x^{2}+1+3x \right )\left ( x^{2}+1+5x \right )

\fn_cm =\left ( x^{2}+3x+1 \right )\left ( x^{2}+5x+1 \right )

উৎপাদকে বিশ্লেষণ

(vi) \fn_cm {\color{Blue} \left ( a-1 \right )x^{2}-x-\left ( a-2 \right )}

সমাধানঃ 

\fn_cm \left ( a-1 \right )x^{2}-x-\left ( a-2 \right )

=ax^{2}-x^{2}-x-a+2

=ax^{2}-x^{2}-x-a+1+1

=ax^{2}-a-x^{2}+1-x+1

=a\left ( x^{2}-1 \right )-\left ( x^{2}-1 \right )-1\left ( x-1 \right )

=a\left ( x+1 \right )\left ( x-1 \right )-\left ( x+1 \right )\left ( x-1 \right )-1\left ( x-1 \right )

[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]

=\left ( x-1 \right )\left [ a\left ( x+1 \right )-\left ( x+1 \right ) -1\right ]

=\left ( x-1 \right )\left ( ax+a-x-1-1 \right )

=\left ( x-1 \right )\left ( ax+a-x-2 \right )

উৎপাদকে বিশ্লেষণ

(vii) \fn_cm {\color{Blue} \left ( a-1 \right )x^{2}+a^{2}xy+\left ( a+1 \right )y^{2}}

সমাধানঃ 

\left ( a-1 \right )x^{2}+a^{2}xy+\left ( a+1 \right )y^{2}

=\left ( a-1 \right )x^{2}+\left [ \left ( a^{2}-1 \right )+1 \right ]xy+\left ( a+1 \right )y^{2}

=\left ( a-1 \right )x^{2}+ \left ( a^{2}-1 \right )xy+xy+\left ( a+1 \right )y^{2}

=\left ( a-1 \right )x\left [ x+\left ( a+1 \right )y \right ]+y\left [ x+\left ( a+1 \right )y \right ]

=\left ( ax-x \right )\left [ x+ay+y \right ]+y\left [ x+ay+y \right ]

=\left ( x+ay+y \right )\left ( ax-x+y \right )

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(viii) \fn_cm {\color{Blue} x^{2}-qx-p^{2}+5pq-6q^{2}}

সমাধানঃ 

\fn_cm x^{2}-qx-p^{2}+5pq-6q^{2}

\fn_cm =x^{2}-qx-\left (p^{2}-5pq+6q^{2} \right )

\fn_cm =x^{2}-qx-\left (p^{2}-3pq-2pq+6q^{2} \right )

\fn_cm =x^{2}-qx-\left [p\left (p-3q \right )-2q\left ( p-3q \right ) \right ]

\fn_cm =x^{2}-qx-\left (p-3q \right )\left ( p-2q \right )

\fn_cm =x^{2}-\left [ \left ( p-2q \right )-\left ( p-3q \right ) \right ]x-\left (p-3q \right )\left ( p-2q \right )

\fn_cm =x^{2}-\left ( p-2q \right )x+\left ( p-3q \right )x-\left (p-3q \right )\left ( p-2q \right )

=x\left ( x-p+2q \right )+\left ( p-3q \right )\left ( x-p+2q \right )

=\left ( x+p-3q \right )\left ( x-p+2q \right )

 

উৎপাদকে বিশ্লেষণ

(ix) \fn_cm {\color{Blue} 2\left ( a^{2}+\frac{1}{a^{2}} \right )-\left ( a-\frac{1}{a} \right )-7}

সমাধানঃ 

\fn_cm {\color{Blue} 2\left ( a^{2}+\frac{1}{a^{2}} \right )-\left ( a-\frac{1}{a} \right )-7}

\fn_cm =2\left [\left ( a-\frac{1}{a} \right )^{2}+2 \right ]-\left ( a-\frac{1}{a} \right )-7

ধরি, {\color{Blue} \left ( a-\frac{1}{a} \right )=x}

\fn_cm =2\left ( x^{2}+2 \right ) -x-7 

\fn_cm =2x^{2}+4-x-7

\fn_cm =2x^{2}-x-3

\fn_cm =2x^{2}-3x+2x-3

=x\left ( 2x-3 \right )+1\left ( 2x-3 \right )

=\left ( 2x-3 \right )\left ( x+1 \right )

x এর মান বসিয়ে পাই,

=\left [ 2\left ( a-\frac{1}{a}-3 \right ) \right ]\left ( a-\frac{1}{a}+1 \right ) 

=\left ( 2a-\frac{2}{a}-3 \right )\left ( a-\frac{1}{a}+1 \right )

 

উৎপাদকে বিশ্লেষণ

(x) \fn_cm {\color{Blue} \left ( x^{2}-x \right )y^{2}+y-\left ( x^{2}+x \right )}

সমাধানঃ 

\fn_cm \left ( x^{2}-x \right )y^{2}+y-\left ( x^{2}+x \right )

\fn_cm =x\left ( x-1 \right )y^{2}+y-x\left ( x+1 \right )

=x\left ( x-1 \right )y^{2}+\left [ x^{2}-\left \{ \left ( x+1 \right )\left ( x-1 \right ) \right \} \right ]y-x\left ( x+1 \right )

=x\left ( x-1 \right )y^{2}+x^{2}y-\left ( x+1 \right )\left ( x-1 \right ) y-x\left ( x+1 \right )

=xy\left [ \left ( x-1 \right )y+x \right ]-\left ( x+1 \right )\left [ \left ( x-1 \right )y+x \right ]

=\left [ \left ( x-1 \right )y+x \right ]\left [xy-\left ( x+1 \right ) \right ]

=\left ( xy-y+x \right )\left ( xy-x-1 \right )

 

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(i) \fn_cm {\color{Blue} a^{2}-b^{2}=11\times 9} এবং a ও b ধনাত্মক পূর্ণসংখ্যা (a > b) হলে 

(a) a = 11, b = 9

(b) a = 33, b = 3

(c) a = 10, b = 1

(d) a = 100, b = 1

সমাধানঃ 

\fn_cm a^{2}-b^{2}=11\times 9

বা, \left ( a+b \right )\left ( a-b \right )=\left ( 10+1 \right )\left ( 10-1 \right )

উভয়পক্ষ তুলনা করে পাই,

a=10,b=1

উত্তরঃ (c) a = 10, b = 1

 

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(ii) যদি  \fn_cm {\color{Blue} \frac{a}{b}+ \frac{b}{a}=1} হয়, তাহলে \fn_cm {\color{Blue} a^{3}+b^{3}} -এর মান 

(a) 1

(b) a

(c) b

(d) 0

সমাধানঃ 

\fn_cm \frac{a}{b}+ \frac{b}{a}=1

বা, \frac{a^{2}+b^{2}}{ab}=1

বা, a^{2}+b^{2}=ab

বা, a^{2}+b^{2}-ab=0

বা, \left ( a+b \right )\left (a^{2}+b^{2}-ab \right )=0\times \left ( a+b \right )

বা, a^{3}+b^{3}=0

উত্তরঃ (d) 0

 

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(iii) \fn_cm {\color{Blue} 25^{3}-75^{3}+50^{3}+3\times 25\times 75\times 50} -এর মান 

(a) 150

(b) 0

(c) 25

(d) 50

সমাধানঃ 

\fn_cm 25^{3}-75^{3}+50^{3}+3\times 25\times 75\times 50

=\left ( 25-75+50 \right )\left [ \left ( 25 \right )^{2}+\left ( -75 \right )^{2}+\left ( 50 \right )^{2}-25\times \left ( -75 \right )-\left ( -75 \right )\times 50-50\times 25 \right ]

{\color{Blue} \left [ \because a^{3}+b^{3}+c^{3}+3abc=\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )\right ]}

= 0

উত্তরঃ (b) 0

 

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(iv) \fn_cm {\color{Blue} a+b+c=0} হলে \fn_cm {\color{Blue} \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}} -এর মান 

(a) 0

(b) 1

(c) −1

(d) 3

সমাধানঃ 

\fn_cm \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}

=\frac{a^{3}+b^{3}+c^{3}}{abc}

=\frac{\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )+3abc}{abc}

=\frac{0\times \left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )}{abc}+\frac{3abc}{abc}

= 0 + 3

= 3

উত্তরঃ (d) 3

 

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(v) \fn_cm {\color{Blue} x^{2}-px+12=\left ( x-3 \right )\left ( x-a \right )} একটি অভেদ হলে, a ও p -এর মান যথাক্রমে 

(a) a = 4, p = 7

(b) a = 7, p = 4

(c) a = 4, p = −7

(d) a = −4, p = 7

সমাধানঃ 

\fn_cm x^{2}-px+12=\left ( x-3 \right )\left ( x-a \right )

বা, \fn_cm x^{2}-px+12=x^{2}-3x-ax+3a

বা, \fn_cm x^{2}-px+12=x^{2}-x\left (3+a \right )+3a

উভয়পক্ষ তুলনা করে পাই,

p=3+a,3a=12

\therefore a=\frac{12}{3}=4

p=3+4=7

উত্তরঃ (a) a = 4, p = 7

 

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(i) \fn_cm {\color{Blue} \frac{\left ( b^{2}-c^{2} \right )^{3}+\left ( c^{2}-a^{2} \right )^{3}+\left ( a^{2}-b^{2} \right )^{3}}{\left ( b-c \right )^{3}+\left ( c-a \right )^{3}+\left ( a-b \right )^{3}}} -এর সরলতম মান লিখি।

সমাধানঃ 

\fn_cm \frac{\left ( b^{2}-c^{2} \right )^{3}+\left ( c^{2}-a^{2} \right )^{3}+\left ( a^{2}-b^{2} \right )^{3}}{\left ( b-c \right )^{3}+\left ( c-a \right )^{3}+\left ( a-b \right )^{3}}

=\frac{\left ( b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2} \right )\left [\left (b^{2}-c^{2} \right )^{2}+\left (c^{2}-a^{2} \right )^{2}+\left (a^{2}-b^{2} \right )^{2}-\left (b^{2}-c^{2} \right )^{2}\left (c^{2}-a^{2} \right )-\left (c^{2}-a^{2} \right )\left (a^{2}-b^{2} \right )-\left (a^{2}-b^{2} \right )\left (b^{2}-c^{2} \right ) \right ]+3\left ( b^{2}-c^{2} \right )\left (c^{2}-a^{2} \right )\left (a^{2}-b^{2} \right )}{\left ( b-c+c-a+a-b \right )\left [ \left ( b-c \right )^{2}+\left ( c-a \right )^{2}+\left ( a-b \right )^{2}-\left ( b-c \right )\left ( c-a \right )-\left ( c-a \right )\left ( a-b \right )-\left ( a-b \right ) \left ( b-c \right )\right ]+3\left ( b-c \right )\left ( c-a \right )\left ( a-b \right )}

{\color{Blue} \left [ \because x^{3}+y^{3}+z^{3}=\left ( x+y+z \right )\left ( x^{2}+y^{2}+z^{2}-xy-yz-zx \right )+3xyz \right ]}

=\frac {3\left ( b^{2}-c^{2} \right )\left (c^{2}-a^{2} \right )\left (a^{2}-b^{2} \right )}{3\left ( b-c \right )\left ( c-a \right )\left ( a-b \right )}

=\left ( b+c \right )\left ( c+a \right )\left ( a+b \right )

 

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(ii) \fn_cm {\color{Blue} a^{3}+b^{3}+c^{3}-3abc=0} এবং \fn_cm {\color{Blue} a+b+c\neq 0} হলে a, b ও c -এর মধ্যে সম্পর্ক লিখি।

সমাধানঃ 

\fn_cm a^{3}+b^{3}+c^{3}-3abc=0

বা, \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )=0

\fn_cm \because a+b+c\neq 0

\therefore \left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )=0

বা, 2\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )=0\times 2

বা, \left ( 2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca \right )=0

বা, a^{2}-2ab+b^{2}+b^{2}-2bc+c^{2}+c^{2}-2ca+a^{2}=0

বা, \left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}=0

আমরা জানি, দুই বা ততোধিক রাশির বর্গের সমষ্টি শূন্য হলে তারা প্রত্যেকে পৃথক পৃথক ভাবে শূন্য হয়। 

\therefore \left ( a-b \right )^{2}=0

বা, a-b=0

বা, a=b

আবার, 

\left ( b-c \right )^{2}=0

বা, b-c=0

বা, b=c

আবারও 

\left ( c-a \right )^{2}=0

বা, c-a=0

বা, c=a

{\color{DarkGreen} \therefore a=b=c} ইহায় a, b ও c -এর মধ্যে সম্পর্ক। 

 

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(iii) \fn_cm {\color{Blue} a^{2}-b^{2}=224} এবং a ও b ঋণাত্মক পূর্ণসংখ্যা হলে (a < b) a ও b -এর মান লিখি।

সমাধানঃ 

\fn_cm a^{2}-b^{2}=224

বা, \left ( a+b \right )\left ( a-b \right )=\left ( -14 \right )\times \left ( -16 \right )

বা, \left ( a+b \right )\left ( a-b \right )=\left ( -15+1 \right )\times \left ( -15-1 \right )

উভয়পক্ষ তুলনা করে পাই,

a=-15,b=-1

উত্তরঃ নির্ণেয় a ও b এর মান যথাক্রমে -15 ও -1

 

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(iv) \fn_cm {\color{Blue} 3x=a+b+c}  হলে, \fn_cm {\color{Blue} \left ( x-a \right )^{3}+\left ( x-b \right )^{3}+\left ( x-c \right )^{3}-3\left ( x-a \right )\left ( x-b \right )\left ( x-c \right )} -এর মান কত লিখি।

সমাধানঃ 

\fn_cm \left ( x-a \right )^{3}+\left ( x-b \right )^{3}+\left ( x-c \right )^{3}-3\left ( x-a \right )\left ( x-b \right )\left ( x-c \right )

ধরি, 

x – a = p , x – b = q ও x – c = r

এখন,

p + q + r = x – a + x – b + x – c

বা, p + q + r = 3x – a – b – c

বা, p + q + r = 3x – 3x  [ যেহেতু, \fn_cm {\color{Blue} 3x=a+b+c} ]

∴ p + q + r = 0

এখন,

\fn_cm \left ( x-a \right )^{3}+\left ( x-b \right )^{3}+\left ( x-c \right )^{3}-3\left ( x-a \right )\left ( x-b \right )\left ( x-c \right )

=p^{3}+q^{3}+r^{3}-3pqr

=\left ( p+q+r \right )\left ( p^{2}+q^{2}+r^{2}-pq-qr-rp \right )+3pqr-3pqr

=0\times \left ( p^{2}+q^{2}+r^{2}-pq-qr-rp \right )

= 0

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(v) \fn_cm {\color{Blue} 2x^{2}+px+6=\left ( 2x-a \right )\left ( x-2 \right )} একটি অভেদ হলে a ও p -এর মান কত লিখি।

সমাধানঃ 

\fn_cm 2x^{2}+px+6=\left ( 2x-a \right )\left ( x-2 \right )

বা, \fn_cm 2x^{2}+px+6=2x^{2}-4x-ax+2a

বা, \fn_cm 2x^{2}+px+6=2x^{2}-\left (4+a \right )x+2a

উভয়পক্ষ তুলনা করে পাই,

p=-\left ( 4+a \right ) এবং 6=2a

\therefore a=\frac{6}{2}=3

এবং 

p=-\left ( 4+a \right ) 

বা, p=-\left ( 4+3 \right ) [ a এর মান বসিয়ে পাই ]

\therefore p=-7

উত্তরঃ নির্ণেয় a ও p এর মান যথাক্রমে 3 ও -7

 

Koshe dekhi 8.5 Class 9

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